A number consist of 2 digit.The digit in tens place exceeds the digit in ones place by 2.The difference of digits is 1/32 of the number.Find the number

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Solution

$\begin{array}{l} Let the digit in unit's place = x \\ Let the digit in ten's place = y \\ The number = 10 y + x \\ A/q \\ y = x + 2 \\ = > y - x = 2 -----1 . \\ y - x = \frac{1}{32} (10 y + x) -----2 . \\ Equating 1 and 2 . \\ \frac{1}{32} (10 y + x) = 2 \\ = > (10 y + x) = 64 \\ The required number = 64 \end{array}$

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A number consist of 2 digit.The digit in tens place exceeds the digit in ones place by 2.The difference of digits is 1/32 of the number.Find the number

Let the digit in unit's place = xLet the digit in ten's place = yThe number = 10y+xA/qy=x+2=>y−x=2 -----1.y−x=132(10y+x) -----2. Equating 1 and 2.132(10y+x)=2=>(10y+x)=64The required number = 64